Question

A random sample of size 4 is taken from the distribution with the probability density function \[ f(x; \theta) = \begin{cases} \frac{2(\theta - x)}{\theta^2}, & 0<x<\theta, \\ 0, & \text{elsewhere}. \end{cases} \] If the observed sample values are 6, 5, 3, 6, then the method of moments estimate (in integer) of the parameter \( \theta \), based on these observations, is ________

Hint:

For the method of moments, equate the sample moments to the population moments and solve for the parameter.

Answer 1

Correct Answer: 15

To find the method of moments estimate of \( \theta \), we first compute the first moment of the distribution. The first moment is the expected value \( E[X] \). The expected value of \( X \) is: \[ E[X] = \int_0^\theta x \frac{2(\theta - x)}{\theta^2} \, dx. \] Simplifying the integral: \[ E[X] = \frac{2}{\theta^2} \int_0^\theta x (\theta - x) \, dx = \frac{2}{\theta^2} \int_0^\theta (\theta x - x^2) \, dx. \] Evaluating the integrals: \[ \int_0^\theta \theta x \, dx = \frac{\theta^2}{2}, \quad \int_0^\theta x^2 \, dx = \frac{\theta^3}{3}. \] Thus: \[ E[X] = \frac{2}{\theta^2} \left( \frac{\theta^3}{2} - \frac{\theta^4}{3} \right) = \frac{\theta}{3}. \] The method of moments estimator equates the sample mean to the population mean. The sample mean of the values 6, 5, 3, 6 is: \[ \bar{X} = \frac{6 + 5 + 3 + 6}{4} = 5. \] Setting \( E[X] = 5 \) gives: \[ \frac{\theta}{3} = 5 \quad \Rightarrow \quad \theta = 15. \] Thus, the method of moments estimate of \( \theta \) is \( \boxed{15} \).